Prove the following assertion: $a<0, b<0,$ then $ab>0$

Question

I am reading Inequalities by Radmila Bulajich Manfrino. I am new to Inequalities, so I don't understand a lot. The above mentioned problem is exercise $1.2(i)$ in the book. There are some properties mentioned before this exercise. They are:

$1.1.1$ Every real number $x$ has one and only one of the following properties: $(i) x = 0 \ (ii) x>0\ (iii) -x>0$

$1.1.2$ If $x>0, y>0 => x+y>0$

$1.1.3$ If $x>0, y>0 => xy>0$

I tried to prove it in the following way:

$a<0 => -a>0$

$b<0 => -b>0$ (from property $1.1.1$)

$(-a)(-b)>0$ (from property $1.1.3$)

So, $ab>0$

I am confused if this is the correct proof. Can I write $(-a)(-b)=ab$, without referring to any properties?

Answer 1

Although you claim you don't 'understand a lot', it is important to have the confidence in yourself that in what matters, you can be 'dead on balls accurate'.

Now what you are trying to prove you've known to be true from an early age, but now you want to be more sophisticated and abstract, using symbols that represent numbers together with some accepted properties.

I don't have your book, but I can wrap this up for you if we can agree on four more properties or 'things'. If the book doesn't explicitly state $\text{(1)}$ thru $\text{(3)}$ as valid properties, I am sure you can derive them within the book's framework.

$$\tag 1 (+1) \times u = u \text{ for any number } u$$ $$\tag 2 -u = (-1) \times u \text{ for any number } u$$ $$\tag 3 (-1) \times (-1) = 1$$ $$\tag 4 \text{Multiplication is both commutative and associative.}$$

You have shown that if both $a$ and $b$ are negative, that the product of $-a$ and $-b$ is positive. But $-a = (-1) \times a$ and $-b = (-1) \times b$, Now using $\text{(4)}$ it can be shown that
for every real numbers $s,t,u$ and $v$,

$$\tag 5 (su) \times (tv) = (st) \times (uv)$$

So,

$$\tag 6 0 \lt (-a)(-b) = ((-1) a) \times ((-1)b) = ((-1)(-1)) \times ab = (+1) \times ab = ab$$

Answer 2

I found the book the OP is using.

Working within the book's framework, the OP can also use the following argument:

Since $a \lt 0$ we can write $a + p = 0$ for some $p \gt 0$.
Since $b \lt 0\,$ we can write $b + q = 0$ for some $q \gt 0$.

Multiplying

$\tag 1 0 = 0 \times 0 = (a+p)(b+q) = ab +aq + pb + pq$

Since $aq = -pq$ and $pb = -pq$, by substituting in $\text{(1)}$,

$\tag 2 0 = ab - pq - pq + pq$

But then $ab = pq$. Using Property $1.1.3$ we see that the product of two negative numbers is positive.